如何获得一个int数字的单独数字？

Question

I have numbers like 1100, 1002, 1022 etc. I would like to have the individual digits, for example for the first number 1100 I want to have 1, 1, 0, 0.

How can I get it in Java?

Answer 1

To do this, you will use the % (mod) operator.

int number; // = some int

while (number > 0) {
    print( number % 10);
    number = number / 10;
}

The mod operator will give you the remainder of doing int division on a number.

So,

10012 % 10 = 2

Because:

10012 / 10 = 1001, remainder 2

Note: As Paul noted, this will give you the numbers in reverse order. You will need to push them onto a stack and pop them off in reverse order.

Code to print the numbers in the correct order:

int number; // = and int
LinkedList<Integer> stack = new LinkedList<Integer>();
while (number > 0) {
    stack.push( number % 10 );
    number = number / 10;
}

while (!stack.isEmpty()) {
    print(stack.pop());
}

Answer 2

Convert it to String and use String#toCharArray() or String#split() .

String number = String.valueOf(someInt);

char[] digits1 = number.toCharArray();
// or:
String[] digits2 = number.split("(?<=.)");

In case you're already on Java 8 and you happen to want to do some aggregate operations on it afterwards, consider using String#chars() to get an IntStream out of it.

IntStream chars = number.chars();

Answer 3

How about this?

public static void printDigits(int num) {
    if(num / 10 > 0) {
        printDigits(num / 10);
    }
    System.out.printf("%d ", num % 10);
}

or instead of printing to the console, we can collect it in an array of integers and then print the array:

public static void main(String[] args) {
    Integer[] digits = getDigits(12345);
    System.out.println(Arrays.toString(digits));
}

public static Integer[] getDigits(int num) {
    List<Integer> digits = new ArrayList<Integer>();
    collectDigits(num, digits);
    return digits.toArray(new Integer[]{});
}

private static void collectDigits(int num, List<Integer> digits) {
    if(num / 10 > 0) {
        collectDigits(num / 10, digits);
    }
    digits.add(num % 10);
}

If you would like to maintain the order of the digits from least significant (index[0]) to most significant (index[n]), the following updated getDigits() is what you need:

/**
 * split an integer into its individual digits
 * NOTE: digits order is maintained - i.e. Least significant digit is at index[0]
 * @param num positive integer
 * @return array of digits
 */
public static Integer[] getDigits(int num) {
    if (num < 0) { return new Integer[0]; }
    List<Integer> digits = new ArrayList<Integer>();
    collectDigits(num, digits);
    Collections.reverse(digits);
    return digits.toArray(new Integer[]{});
}

Answer 4

I haven't seen anybody use this method, but it worked for me and is short and sweet:

int num = 5542;
String number = String.valueOf(num);
for(int i = 0; i < number.length(); i++) {
    int j = Character.digit(number.charAt(i), 10);
    System.out.println("digit: " + j);
}

This will output:

digit: 5
digit: 5
digit: 4
digit: 2

Answer 5

I noticed that there are few example of using Java 8 stream to solve your problem but I think that this is the simplest one:

int[] intTab = String.valueOf(number).chars().map(Character::getNumericValue).toArray();

To be clear: You use String.valueOf(number) to convert int to String, then chars() method to get an IntStream (each char from your string is now an Ascii number), then you need to run map() method to get a numeric values of the Ascii number. At the end you use toArray() method to change your stream into an int[] array.

Answer 6

I see all the answer are ugly and not very clean.

I suggest you use a little bit of recursion to solve your problem. This post is very old, but it might be helpful to future coders.

public static void recursion(int number) {
    if(number > 0) {
        recursion(number/10);
        System.out.printf("%d   ", (number%10));
    }
}

Output:

Input: 12345

Output: 1   2   3   4   5 

Answer 7

simple solution

public static void main(String[] args) {
    int v = 12345;
    while (v > 0){
        System.out.println(v % 10);
        v /= 10;
    }
}

Answer 8

// could be any num this is a randomly generated one
int num = (int) (Math.random() * 1000);

// this will return each number to a int variable
int num1 = num % 10;
int num2 = num / 10 % 10;
int num3 = num /100 % 10;

// you could continue this pattern for 4,5,6 digit numbers
// dont need to print you could then use the new int values man other ways
System.out.print(num1);
System.out.print("\n" + num2);
System.out.print("\n" + num3);

Answer 9

Since I don't see a method on this question which uses Java 8, I'll throw this in. Assuming that you're starting with a String and want to get a List<Integer> , then you can stream the elements like so.

List<Integer> digits = digitsInString.chars()
        .map(Character::getNumericValue)
        .boxed()
        .collect(Collectors.toList());

This gets the characters in the String as a IntStream , maps those integer representations of characters to a numeric value, boxes them, and then collects them into a list.

Answer 10

Easier way I think is to convert the number to string and use substring to extract and then convert to integer.

Something like this:

int digits1 =Integer.parseInt( String.valueOf(201432014).substring(0,4));
    System.out.println("digits are: "+digits1);

ouput is 2014

Answer 11

Java 9 introduced a new Stream.iterate method which can be used to generate a stream and stop at a certain condition. This can be used to get all the digits in the number, using the modulo approach.

int[] a = IntStream.iterate(123400, i -> i > 0, i -> i / 10).map(i -> i % 10).toArray();

Note that this will get the digits in reverse order, but that can be solved either by looping through the array backwards (sadly reversing an array is not that simple ), or by creating another stream:

int[] b = IntStream.iterate(a.length - 1, i -> i >= 0, i -> i - 1).map(i -> a[i]).toArray();

or

int[] b = IntStream.rangeClosed(1, a.length).map(i -> a[a.length - i]).toArray();

As an example, this code:

int[] a = IntStream.iterate(123400, i -> i > 0, i -> i / 10).map(i -> i % 10).toArray();
int[] b = IntStream.iterate(a.length - 1, i -> i >= 0, i -> i - 1).map(i -> a[i]).toArray();
System.out.println(Arrays.toString(a));
System.out.println(Arrays.toString(b));

Will print:

[0, 0, 4, 3, 2, 1]
[1, 2, 3, 4, 0, 0]

Answer 12

I wrote a program that demonstrates how to separate the digits of an integer using a more simple and understandable approach that does not involve arrays, recursions, and all that fancy schmancy. Here is my code:

int year = sc.nextInt(), temp = year, count = 0;

while (temp>0)
{
  count++;
  temp = temp / 10;
}

double num = Math.pow(10, count-1);
int i = (int)num;

for (;i>0;i/=10)
{
  System.out.println(year/i%10);
}

Suppose your input is the integer 123 , the resulting output will be as follows:

1
2
3

Answer 13

Here is my answer, I did it for myself and I hope it's simple enough for those who don't want to use the String approach or need a more math-y solution:

public static void reverseNumber2(int number) {

    int residual=0;
    residual=number%10;
    System.out.println(residual);

    while (residual!=number)  {
          number=(number-residual)/10;
          residual=number%10;
          System.out.println(residual);
    }
}

So I just get the units, print them out, substract them from the number, then divide that number by 10 - which is always without any floating stuff, since units are gone, repeat.

Answer 14

Why don't you do:

String number = String.valueOf(input);
char[] digits = number.toCharArray();

Answer 15

Java 8 解决方案，用于从作为字符串的整数获取数字为 int[]：

int[] digits = intAsString.chars().map(i -> i - '0').toArray();

Answer 16

chars()和codePoints()都不是——另一个 lambda

String number = Integer.toString( 1100 );

IntStream.range( 0, number.length() ).map( i -> Character.digit( number.codePointAt( i ), 10 ) ).toArray();  // [1, 1, 0, 0]

Answer 17

This uses the modulo 10 method to figure out each digit in a number greater than 0, then this will reverse the order of the array. This is assuming you are not using "0" as a starting digit.

This is modified to take in user input. This array is originally inserted backwards, so I had to use the Collections.reverse() call to put it back into the user's order.

    Scanner scanNumber = new Scanner(System.in);
    int userNum = scanNumber.nextInt(); // user's number

    // divides each digit into its own element within an array
    List<Integer> checkUserNum = new ArrayList<Integer>();
    while(userNum > 0) {
        checkUserNum.add(userNum % 10);
        userNum /= 10;
    }

    Collections.reverse(checkUserNum); // reverses the order of the array

    System.out.print(checkUserNum);

Answer 18

int number = 12344444; // or it Could be any valid number

int temp = 0;
int divider = 1;

for(int i =1; i< String.valueOf(number).length();i++)
 {

    divider = divider * 10;

}

while (divider >0) {

    temp = number / divider;
    number = number % divider;
    System.out.print(temp +" ");
    divider = divider/10;
}

Answer 19

Just to build on the subject, here's how to confirm that the number is a palindromic integer in Java:

public static boolean isPalindrome(int input) {
List<Integer> intArr = new ArrayList();
int procInt = input;

int i = 0;
while(procInt > 0) {
    intArr.add(procInt%10);
    procInt = procInt/10;
    i++;
}

int y = 0;
int tmp = 0;
int count = 0;
for(int j:intArr) {
    if(j == 0 && count == 0) {
    break;
    }

    tmp = j + (tmp*10);
    count++;
}

if(input != tmp)
    return false;

return true;
}

I'm sure I can simplify this algo further. Yet, this is where I am. And it has worked under all of my test cases.

I hope this helps someone.

Answer 20

Try this:

int num= 4321
int first  =  num % 10;
int second =  ( num - first ) % 100 / 10;
int third  =  ( num - first - second ) % 1000 / 100;
int fourth =  ( num - first - second - third ) % 10000 / 1000;

You will get first = 1, second = 2, third = 3 and fourth = 4 ....

Answer 21

Integer.toString(1100) gives you the integer as a string. Integer.toString(1100).getBytes() to get an array of bytes of the individual digits.

Edit:

You can convert the character digits into numeric digits, thus:

  String string = Integer.toString(1234);
  int[] digits = new int[string.length()];

  for(int i = 0; i<string.length(); ++i){
    digits[i] = Integer.parseInt(string.substring(i, i+1));
  }
  System.out.println("digits:" + Arrays.toString(digits));

Answer 22

public int[] getDigitsOfANumber(int number) {
    String numStr = String.valueOf(number);
    int retArr[] = new int[numStr.length()];

    for (int i = 0; i < numStr.length(); i++) {
        char c = numStr.charAt(i);
        int digit = c;
        int zero = (char) '0';
        retArr[i] = digit - zero;

    }
    return retArr;
}

Answer 23

Something like this will return the char[] :

public static char[] getTheDigits(int value){
    String str = "";
    int number = value;
    int digit = 0;
    while(number>0){
        digit = number%10;
        str = str + digit;
        System.out.println("Digit:" + digit);
        number = number/10;     

    }
    return str.toCharArray();
}

Answer 24

As a noob, my answer would be:

String number = String.valueOf(ScannerObjectName.nextInt()); 
int[] digits = new int[number.length()]; 
for (int i = 0 ; i < number.length() ; i++)
    int[i] = Integer.parseInt(digits.substring(i,i+1))

Now all the digits are contained in the "digits" array.

Answer 25

if digit is meant to be a Character

String numstr = Integer.toString( 123 );
Pattern.compile( "" ).splitAsStream( numstr ).map(
  s -> s.charAt( 0 ) ).toArray( Character[]::new );  // [1, 2, 3]

and the following works correctly
numstr = "000123" gets [0, 0, 0, 1, 2, 3]
numstr = "-123" gets [-, 1, 2, 3]

Answer 26

Try this one.

const check = (num) => {
  let temp = num
  let result = []
  while(temp > 0){
    let a = temp%10;
    result.push(a);
    temp = (temp-a)/10;
  }
  return result;
}

check(98) //[ 8, 9 ]

Answer 27

使用 LINQ 的 .NET 解决方案。

List<int> numbers = number.ToString().Select(x => x - 48).ToList();

Answer 28

I think this will be the most useful way to get digits:

public int[] getDigitsOf(int num)
{        
    int digitCount = Integer.toString(num).length();

    if (num < 0) 
        digitCount--;           

    int[] result = new int[digitCount];

    while (digitCount-- >0) {
        result[digitCount] = num % 10;
        num /= 10;
    }        
    return result;
}

Then you can get digits in a simple way:

int number = 12345;
int[] digits = getDigitsOf(number);

for (int i = 0; i < digits.length; i++) {
    System.out.println(digits[i]);
}

or more simply:

int number = 12345;
for (int i = 0; i < getDigitsOf(number).length; i++) {
    System.out.println(  getDigitsOf(number)[i]  );
}

Notice the last method calls getDigitsOf method too much time. So it will be slower. You should create an int array and then call the getDigitsOf method once, just like in second code block.

In the following code, you can reverse to process. This code puts all digits together to make the number:

public int digitsToInt(int[] digits)
{
    int digitCount = digits.length;
    int result = 0;

    for (int i = 0; i < digitCount; i++) {
        result = result * 10;
        result += digits[i];
    }

    return result;
}

Both methods I have provided works for negative numbers too.

Answer 29

see bellow my proposal with comments

          int size=i.toString().length(); // the length of the integer (i) we need to split;
           ArrayList<Integer> li = new ArrayList<Integer>(); // an ArrayList in whcih to store the resulting digits

        Boolean b=true; // control variable for the loop in which we will reatrive step by step the digits
        String number="1"; // here we will add the leading zero depending on the size of i
        int temp;  // the resulting digit will be kept by this temp variable

    for (int j=0; j<size; j++){
                        number=number.concat("0");
                    }

Integer multi = Integer.valueOf(number); // the variable used for dividing step by step the number we received 
                while(b){

                    multi=multi/10;
                    temp=i/(multi);
                    li.add(temp);
                    i=i%(multi);
                                        if(i==0){
                                        b=false;
                                        }


                }

                for(Integer in: li){
                    System.out.print(in.intValue()+ " ");
                }

Answer 30

import java.util.Scanner;

class  Test 
{  
    public static void main(String[] args)   
    {  
        Scanner sc = new Scanner(System.in); 


    int num=sc.nextInt(); 
    System.out.println("Enter a number (-1 to end):"+num);
    int result=0;
    int i=0;
    while(true) 
    { 
      int n=num%10;
      if(n==-1){
        break;
      }
      i++;
      System.out.println("Digit"+i+" = "+n);
      result=result*10+n;
      num=num/10; 


      if(num==0) 
      { 
        break; 
      } 
    }
    }
}

Answer 31

in Java, this is how you can separate digits from numbers and store them in an Array.

public static void main(String[] args) {
        System.out.println("Digits Array:: "+Arrays.toString(getNumberArr(1100)));
}

private static Integer[] getNumberArr(int number) {
    //will get the total number of digits in the number
    int temp = number;
    int counter = 0;

    while (temp > 0) {
        temp /= 10;
        counter++;
    }
    //reset the temp
    temp = number;

    // make an array
    int modulo;     //modulo is equivalent to single digit of the number.
    Integer[] numberArr = new Integer[counter];
    for (int i = counter - 1; i >= 0; i--) {
        modulo = temp % 10;
        numberArr[i] = modulo;  
        temp /= 10;
    }

    return numberArr;
}

Output:

Digits Array:: [1, 1, 0, 0]

Answer 32

import java.util.Scanner;

public class SeparatingDigits {

    public static void main( String[] args )
    {

        System.out.print( "Enter the digit to print separately :- ");
        Scanner scan = new Scanner( System.in );

        int element1 = scan.nextInt();
        int divider;

        if( ( element1 > 9999 ) && ( element1 <= 99999 ) )
        {
            divider = 10000;
        }
        else if( ( element1 > 999 ) && ( element1 <= 9999 ) )
        {
            divider = 1000;
        }
        else if ( ( element1 > 99) && ( element1 <= 999 ) )
        {
            divider = 100;
        }
        else if( ( element1 > 9 ) && ( element1 <= 99 ) )
        {
            divider = 10;
        }
        else 
        {
            divider = 1;
        }

        quotientFinder( element1, divider );




    }

     public static void quotientFinder( int elementValue, int dividerValue )
     {
         for( int count = 1;  dividerValue != 0; count++)
         {
            int quotientValue = elementValue / dividerValue ;
            elementValue = elementValue % dividerValue ;
            System.out.printf( "%d  ", quotientValue );

            dividerValue /= 10;

         }
     }
    }

Without using arrays and Strings . ( digits 1-99999 )

output :

Enter the digit to print separately :- 12345

1 2 3 4 5